IMC 2005 · Day 2 · P8

Yes, it is even enough to assume that $f^2$ and $f^3$ are polynomials.

Let $p = f^2$ and $q = f^3$. Write these polynomials in the form of $$p = a \cdot p_1^{a_1} \cdot \dots \cdot p_k^{a_k}, \qquad q = b \cdot q_1^{b_1} \cdot \dots \cdot q_l^{b_l},$$ where $a, b \in \mathbb{R}$, $a_1, \dots, a_k, b_1, \dots b_l$ are positive integers and $p_1, \dots, p_k, q_1, \dots, q_l$ are irreducible polynomials with leading coefficients 1. For $p^3 = q^2$ and the factorisation of $p^3 = q^2$ is unique we get that $a^3 = b^2$, $k = l$ and for some $(i_1, \dots, i_k)$ permutation of $(1, \dots, k)$ we have $p_1 = q_{i_1}, \dots, p_k = q_{i_k}$ and $3 a_1 = 2 b_{i_1}, \dots, 3 a_k = 2 b_{i_k}$. Hence $b_1, \dots, b_l$ are divisible by 3 let

$r = b^{1/3} \cdot q_1^{b_1/3} \cdot \dots \cdot q_l^{b_l/3}$ be a polynomial. Since $r^3 = q = f^3$ we have $f = r$.

Let $\frac{p}{q}$ be the simplest form of the rational function $\frac{f^3}{f^2}$. Then the simplest form of its square is $\frac{p^2}{q^2}$. On the other hand $\frac{p^2}{q^2} = \left( \frac{f^3}{f^2} \right)^2 = f^2$ is a polynomial therefore $q$ must be a constant and so $f = \frac{f^3}{f^2} = \frac{p}{q}$ is a polynomial.